An empty tank be filled with an inlet pipe ‘A’ in 42 minutes. after 12 minutes an outlet pipe ‘B’
is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe
‘C’ opened into the same tank, which can fill the tank in 35 minutes and the tank is filled find the
time taken to fill the tank?

• let the tank be filled in n minutes after opening tap C
  now total part of tank filled=[(n/42)-(n/30)+(n/35)]
  now
  (12/42)-(6/30)+(6/42)+[(n/42)-(n/30)+(n/35)=1
  solve for n we get n=40.5 minutes
  therefore
  total time= 40.5min + 12min(when A was opened) + 6 min(when B was opened)= 58.5 min ans....
  
  
• 12 years agoHelpfull: Yes(34) No(11)
• sorry i had done a mistake in the solution.
  12[1/42]+6[1/42-1/30]+t[1/42-1/30+1/35]=1
  t=40.5
  After getting the t value, the total time is t= 40.5+12+6.=58.5 mins
• 12 years agoHelpfull: Yes(14) No(4)
• Let C has been opening for t min.
  then the equation will be
  =>((12+6+t)/42)+(t/35)-((6+t)/30)=1
  => 30+60+5t+6t-42-7t=210
  =>48+4t=210
  =>4t=210
  =>t=40.5 min
  Total time will be 12+6+40.5=58.5 min
• 11 years agoHelpfull: Yes(6) No(1)
• time to fill the tank=(1/42)-(1/30)+(1/35)=2/105 i.e,105/2=52.5
  subtract 12-6=6
  add it to the 52.5 we get the total time =58.5min
• 10 years agoHelpfull: Yes(4) No(1)
• Part of tank filled in 1min=1/42
  After 12 min part of filled=12*(1/42)
  Now tank emptied in 1 min=1/30
  After 6 min part emptied=6*(1/30)
  so,
  (12/42+1/35-6/30)=4/35
  
  time taken to fill tank=35/4 minutes
• 12 years agoHelpfull: Yes(3) No(14)
• its 58.5 min
• 12 years agoHelpfull: Yes(2) No(1)
• x/42-(x+12)/30+(x+18)/35 = 1 take lcm
  5x-(x+12)7+(x+18)6 = 210
  4x = 234
  x = 58.5
• 9 years agoHelpfull: Yes(1) No(0)
• 12[1/42]+6[1/42-1/30]+t[1/42-1/30+1/35]=1
  t=40.5
  total tym=40.5+35/8=44.8
• 12 years agoHelpfull: Yes(0) No(7)